Question

A train is moving on straight track with velocity $v_0=13.5m{s}^{-1}$. To stop the train at a particular station, the driver applies brakes at $t=0$, which is caused of a retardation proportional to the velocity of the train. The speed of train reduces $50\%$ in the first $t_0=4$ in $2s$. The velocity of train (in $m{s}^{-1}$ ) at $t=4s$ (Given, $e=2.7$ ) is_______.

Answer 1

Answer: 5

$\because a=-kv$ or $\frac{dv}{dt}=-kv$
Or $\underset{v_0}{\overset{v}{\int }}\frac{dv}{v}=-k\underset{0}{\overset{t}{\int }}dt$
or $[\ln v{]}_{v_0}^v=-kt$
Or In $v-\ln v_0=-kt$
or $\ln \frac{v}{v_0}=-kt$
Or $\frac{v}{v_0}=e^{-kt}$
For $t=t_0,v=v_0-v_0\times \frac{50}{100}=\frac{v_0}{2}$
$\therefore \frac{v_0}{2}=v_0{e}^{-kt}$
Or $v=v_0{e}^{\frac{-\ln 2}{t_0}t}$
or $\ln v=\ln v_0-\frac{\ln 2}{t_0}t$
Or $\ln \frac{v}{v_0}=-\frac{\ln 2}{4\ln 2}t$
or $\ln \frac{v}{v_0}=-\frac{t}{4}$
or $\frac{v}{v_0}=e^{-\frac{t}{4}}$
$\therefore v=v_0{e}^{-\frac{t}{4}}=13.5e^{-\frac{4}{4}}$
$=\frac{13.5}{e}=\frac{13.5}{2.7}=5m{s}^{-1}$